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Board exam practice paper
CBSE · Class 10 · Mathematics (Standard) (041)

Mathematics (Standard): Board Exam Practice Paper

Time allowed: 3 hours · Maximum marks: 80 · AW-10M-BP-70027

Name: Date: Score: ____ / 80
  1. This paper has 38 questions. All questions are compulsory.
  2. It is in five sections: A, B, C, D and E.
  3. Section A has 20 multiple-choice questions (18 MCQs and 2 assertion–reason questions) of 1 mark each.
  4. Section B has 5 very short answer questions of 2 marks each.
  5. Section C has 6 short answer questions of 3 marks each.
  6. Section D has 4 long answer questions of 5 marks each.
  7. Section E has 3 case-based questions of 4 marks each, with sub-parts of 1, 1 and 2 marks.
  8. Some questions have an internal choice (marked OR): answer only one of them.
  9. Draw neat figures wherever needed. Take π = 22/7 wherever needed, unless the question says otherwise.
  10. Use of a calculator is not allowed.
Section A: 20 questions of 1 mark each
1.
[1]The LCM of 36 and 54 is:
  1. 1944
  2. 90
  3. 18
  4. 108
2.
[1]If −7 is a zero of the polynomial x2 + kx + 21, then k is:
  1. 10
  2. −3
  3. −10
  4. −7
3.
[1]For what value of k does 4x + 4y = 8 and 12x + 12y = k have infinitely many solutions?
  1. k = −24
  2. k = 8
  3. k = 24
  4. k = 11
4.
[1]The roots of x2 − x − 12 = 0 are:
  1. real and equal
  2. real and distinct
  3. always zero
  4. not real (no real roots)
5.
[1]The sum of the first 26 terms of 5, 3, 1, … is:
  1. −1040
  2. −567
  3. 78
  4. −520
6.
[1]If two angles of one triangle equal two angles of another, the triangles are similar by:
  1. RHS
  2. SSS
  3. SAS
  4. AA
7.
[1]The distance of (−15, −8) from the origin is:
  1. 7 units
  2. 17 units
  3. 289 units
  4. 23 units
8.
[1]The distance between (0, 0) and (3, 4) is:
  1. 7
  2. 5
  3. 25
  4. 1
9.
[1]The value of tan 45° tan 30° + sin 60° sin 60° is:
  1. −3/4 + √3/3
  2. 1/√3
  3. 3/4 + √3/3
  4. 3/2 + 2√3/3
10.
[1]For an acute angle A, sin A is always:
  1. between 0 and 1
  2. greater than 1
  3. negative
  4. equal to 1
11.
[1]A vertical pole 21 m tall casts a shadow 21√3 m long. The angle of elevation of the sun is:
  1. 90°
  2. 30°
  3. 45°
  4. 60°
12.
[1]PA and PB are tangents from P to a circle with centre O. If ∠APB = 50°, then ∠AOB is:
  1. 65°
  2. 130°
  3. 50°
  4. 310°
13.
[1]In one complete turn, a wheel of radius r moves forward by:
  1. πr
  2. 4πr
  3. πr2
  4. 2πr
14.
[1]For the data below, the modal class is:
Class: 20–25, 25–30, 30–35, 35–40, 40–45
Frequency: 3, 18, 13, 20, 14
  1. 30–35
  2. 20–25
  3. 25–30
  4. 35–40
15.
[1]The mean of 32, 89, 36, 69, 75, 57, 46 is:
  1. 59.71
  2. 50.5
  3. 57.71
  4. 404
16.
[1]For the data below, the median class is:
Class: 10–20, 20–30, 30–40, 40–50, 50–60, 60–70
Frequency: 6, 4, 19, 10, 13, 12
  1. 20–30
  2. 30–40
  3. 40–50
  4. 10–20
17.
[1]A card is drawn from a well-shuffled deck of 52 cards. The probability that it is a face card is:
  1. 3/13
  2. 1/4
  3. 10/13
  4. 3/26
18.
[1]A die is thrown once. The probability of getting a prime number is:
  1. 2/3
  2. 1/4
  3. 1/2
  4. 1/3
19.
[1]Assertion (A): If HCF(14, 22) = 2, then LCM(14, 22) = 154.
Reason (R): For any two positive integers a and b, HCF(a, b) × LCM(a, b) = a + b.
Choose the correct option.
  1. Both A and R are true, and R is the correct explanation of A.
  2. Both A and R are true, but R is not the correct explanation of A.
  3. A is true, but R is false.
  4. A is false, but R is true.
20.
[1]Assertion (A): 216 is a term of the AP 9, 15, 21, ….
Reason (R): A number x is a term of an AP with first term a and common difference d exactly when (x − a)/d is a whole number.
Choose the correct option.
  1. Both A and R are true, and R is the correct explanation of A.
  2. Both A and R are true, but R is not the correct explanation of A.
  3. A is true, but R is false.
  4. A is false, but R is true.
Section B: 5 questions of 2 marks each
21.
[2]Express 11,979 as a product of its prime factors.
22.
[2]Solve: 71x + 69y = 978 and 69x + 71y = 982
23.
[2]In △ABC, DE ∥ BC. If AD = x, DB = x − 2, AE = x + 4 and EC = x + 1 (in cm), find x.
OR
[2]ABCD circumscribes a circle. AB = 18 cm, BC = 20 cm and CD = 22 cm. Find DA and the perimeter of ABCD.
24.
[2]Find the perimeter of a sector of angle 100° in a circle of radius 14 cm (π = 22/7).
OR
[2]Find the area of the major sector of a circle of radius 21 cm when the minor sector has an angle of 135° (π = 22/7).
25.
[2]A jar has 50 marbles. The probability of drawing a red marble is 2/5. How many red marbles are there?
Section C: 6 questions of 3 marks each
26.
[3]The floor of a hall measures 105 cm by 90 cm. It is to be paved with square tiles, all of the same size, with no tile cut. Find the side of the largest tile that can be used, and the number of such tiles needed.
27.
[3]If α and β are the zeroes of the quadratic polynomial 2x2 − x − 8, find the value of (α − β)2.
28.
[3]△ABC ~ △PQR. The areas of △ABC and △PQR are 8 cm2 and 2 cm2 respectively. If BC = 10 cm, find QR.
29.
[3]PA and PB are tangents drawn from an external point P to a circle with centre O and radius 18 cm. If PA = 80 cm, find OP and the perimeter of the quadrilateral OAPB.
OR
[3]A circle is inscribed in △ABC, touching the sides BC, CA and AB at D, E and F respectively. If AB = 19 cm, BC = 16 cm and CA = 17 cm, find the lengths of AF, BD and CE.
30.
[3]A chord of a circular mirror of radius 20 cm subtends an angle of 90° at the centre. Find the area of the corresponding (i) minor segment and (ii) major segment. (Use π = 3.14.)
31.
[3]The following table gives the distances (in km) travelled by the members of a cycling club on a Sunday. Find the mode of the data.
Class interval (distance in km) and frequency (number of members): 10–15: 16, 15–20: 4, 20–25: 12, 25–30: 10, 30–35: 18, 35–40: 10.
OR
[3]The following table gives the marks obtained by the students of a class in a test. Find the mean using the direct method.
Class interval (marks) and frequency (number of students): 0–10: 19, 10–20: 15, 20–30: 12, 30–40: 10, 40–50: 19, 50–60: 5.
Section D: 4 questions of 5 marks each
32.
[5]Find two consecutive even positive integers, the sum of whose squares is 2452.
OR
[5]Taxi charges in a city are made up of a fixed charge together with a charge for each kilometre travelled. For a journey of 7 km the charge is Rs 159, and for a journey of 30 km the charge is Rs 550. Find the fixed charge and the charge per kilometre. How much does a person have to pay for a journey of 18 km?
33.
[5]Isha saves Rs 220 in the first month of a year, and in every month after that saves Rs 45 more than in the month before. In how many months will Isha's total savings be Rs 7,175? How much does Isha save in the last of these months?
34.
[5]ZTWD is a trapezium with ZT ∥ DW. L and G are points on the non-parallel sides ZD and TW respectively such that LG ∥ ZT. Prove that ZL/LD = TG/GW. Hence, if ZL = 9 cm, LD = 15 cm and TW = 24 cm, find TG and GW.
35.
[5]An observer at the top of a lighthouse 66 m above sea level watches a ship sailing straight towards the lighthouse. The angle of depression of the ship changes from 30° to 45° during the observation. Find the distance travelled by the ship during this time, and its distance from the lighthouse at the second observation. (Take √3 = 1.73.)
OR
[5]From the top of a building 24 m high, the angle of elevation of the top of a chimney is 45° and the angle of depression of its foot is 30°. The building and the chimney stand on the same level ground. Find the height of the chimney and the distance between the building and the chimney. (Take √3 = 1.73.)
Section E: 3 case-based questions of 4 marks each
36.
A town map
Arjun draws a map of the town on a coordinate grid, with the town hall at the origin and each unit standing for 1 km. The house H is at (−10, −9), the library S is at (−12, −5) and the post office M is at (−15, 1). A straight road runs from H to M, and the main highway of the town runs along the x-axis.
(i)
[1]How far is the library from the town hall?
(ii)
[1]A bus stop is to be built midway between H and M. Find its coordinates.
(iii)
[2]Using the distance formula, show that H, S and M lie on one straight line.
OR
[2]The highway (the x-axis) crosses the road HM. Find the ratio in which it divides HM, and the point where it crosses.
37.
A tower seen from a building
Kabir stands on the flat roof of a building 30 m tall and looks at a mobile tower across the street. The angle of elevation of the top of the tower from the roof is 30°, and the angle of depression of its foot is 60°. The building and the tower stand on the same level ground. Ignore Kabir's own height.
(i)
[1]How far is the building from the tower?
(ii)
[1]How much taller is the tower than the building?
(iii)
[2]Find the height of the tower. (Take √3 = 1.73.)
OR
[2]Find the distance from Kabir to the foot of the tower along the line of sight. (Take √3 = 1.73.)
38.
A gas cylinder
A gas cylinder used in a laboratory is in the shape of a cylinder with a hemisphere at each end. Its total length is 92 cm and its diameter is 42 cm. Kabir works out its surface area and its capacity. (Use π = 22/7.)
(i)
[1]Find the length of the cylindrical part.
(ii)
[1]Find the curved surface area of the cylindrical part.
(iii)
[2]Find the total surface area of the solid.
OR
[2]Find the volume of the solid.
AmbifulWorksheets
Marking scheme
CBSE · Class 10 · Mathematics (Standard) (041)

Marking scheme

Mathematics (Standard): Board Exam Practice Paper · AW-10M-BP-70027 · For the parent or teacher: award the marks shown for each step, even when the final answer is wrong.

QAnswer and marks for each stepMarks
Section A: 20 questions of 1 mark each
1(d) 108
LCM = 18 × 2 × 3 = 108
1
2(a) 10
(−7)2 + k × (−7) + 21 = 0, so k = 10
1
3(c) k = 24
the second equation must be 3 times the first
1
4(b) real and distinct
b2 − 4ac = 49 > 0
1
5(d) −520
26/2 × (2 × 5 + 25 × (−2)) = −520
1
6(d) AA1
7(b) 17 units
√(225 + 64) = 17
1
8(b) 5
√(9 + 16) = 5
1
9(c) 3/4 + √3/3
1/√3 + 3/4
1
10(a) between 0 and 1
the opposite side is shorter than the hypotenuse
1
11(b) 30°
tan θ = 21 ÷ 21√3, so θ = 30°
1
12(b) 130°
∠OAP = ∠OBP = 90°, so ∠AOB = 360° − 180° − 50° = 130°
1
13(d) 2πr
its circumference
1
14(d) 35–40
the highest frequency is 20
1
15(c) 57.71
404 ÷ 7 = 57.71
1
16(c) 40–50
n = 64, n/2 = 32; cumulative frequencies 6, 10, 29, 39, 52, 64
1
17(a) 3/13
12 such cards out of 52
1
18(c) 1/2
3 favourable outcomes out of 6
1
19(c) A is true, but R is false.
A is true: LCM = 14 × 22 ÷ 2 = 154. R is false.
1
20(d) A is false, but R is true.
A is false: (216 − 9) ÷ 6 = 69/2, which is not a whole number. R is true.
1
Section B: 5 questions of 2 marks each
2132 × 113
  1. divide by primes in turn: 32 × 113 (2)
2
22x = 6, y = 8
  1. add: x + y = 14; subtract: x − y = −2 (2)
2
23x = 8
  1. AD/DB = AE/EC: (x)(x + 1) = (x − 2)(x + 4), which gives x = 8 (2)
2
23 ORDA = 20 cm; perimeter 80 cm
  1. AB + CD = BC + DA; perimeter = 2(AB + CD) (2)
2
2452 4/9 cm
  1. 2r + arc = 28 + 24 4/9 (2)
2
24 OR866 1/4 cm2
  1. major sector angle = 360° − 135° = 225° (2)
2
2520 red marbles
  1. 2/5 × 50 = 20 (2)
2
Section C: 6 questions of 3 marks each
26Side of the largest tile = HCF(105, 90) = 15 cm; number of tiles = (105 ÷ 15) × (90 ÷ 15) = 7 × 6 = 42.
  1. Prime factorisation: 105 = 3 × 5 × 7, 90 = 2 × 32 × 5 (1)
  2. Side of the largest tile = HCF = 15 cm (1)
  3. Number of tiles = 7 × 6 = 42 (1)
3
27α + β = 1/2 and αβ = −4. (α − β)2 = (α + β)2 − 4αβ = 65/4.
  1. α + β = 1/2 and αβ = −4 (1)
  2. Writing (α − β)2 = (α + β)2 − 4αβ (1)
  3. Value = 65/4 (1)
3
28QR = 5 cm
  1. ar(ABC)/ar(PQR) = (BC/QR)2, for similar triangles (1)
  2. Substituting the given values (1)
  3. QR = 5 cm (1)
3
29OP = 82 cm; perimeter of OAPB = 196 cm
  1. OA ⟂ PA (radius ⟂ tangent), so △OAP is right-angled at A (1)
  2. OP = 82 cm (1)
  3. PB = PA, perimeter = 196 cm (1)
3
29 ORAF = 10 cm, BD = 9 cm, CE = 7 cm
  1. Tangents from an external point are equal: AF = AE, BD = BF, CD = CE (1)
  2. Equations formed, AF + BD + CE = 26 (1)
  3. AF = 10 cm, BD = 9 cm, CE = 7 cm (1)
3
30Minor segment = 114 cm2; major segment = 1,142 cm2
  1. Area of the sector = (90/360) × 3.14 × 202 = 314 cm2 (1)
  2. area of the right triangle = (1/2) × 20 × 20 = 200 cm2; minor segment = 314 − 200 = 114 cm2 (1)
  3. Major segment = 3.14 × 202 − 114 = 1,142 cm2 (1)
3
31Mode = 32.5
  1. Modal class = 30–35 (highest frequency 18); l = 30, h = 5, f1 = 18, f0 = 10, f2 = 10 (1)
  2. Mode = l + (f1 − f0)/(2f1 − f0 − f2) × h = 30 + (8/16) × 5 (1)
  3. Mode = 32.5 (1)
3
31 ORMean = 26.25
  1. Class marks xi: 5, 15, 25, 35, 45, 55; fixi: 95, 225, 300, 350, 855, 275 (1)
  2. Σfi = 80, Σfixi = 2,100 (1)
  3. Mean = Σfixi/Σfi = 2,100/80 = 26.25 (1)
3
Section D: 4 questions of 5 marks each
32Let the integers be x and x + 2: x2 + (x + 2)2 = 2452, that is, 2x2 + 4x − 2448 = 0, so x2 + 2x − 1224 = 0, so (x − 34)(x + 36) = 0, giving x = 34 or x = −36. The integers are positive, so x = −36 is rejected. The integers are 34 and 36.
  1. Forming the equation: Let the integers be x and x + 2: x2 + (x + 2)2 = 2452, that is, 2x2 + 4x − 2448 = 0 (2)
  2. Solving: x = 34 or x = −36 (2)
  3. Rejecting x = −36: the integers are 34 and 36 (1)
5
32 ORFixed charge = Rs 40; charge per kilometre = Rs 17. For 18 km: Rs 40 + 18 × Rs 17 = Rs 346.
  1. Let the fixed charge be Rs x and the charge per kilometre be Rs y: x + 7y = 159 and x + 30y = 550 (2)
  2. Subtracting: 23y = 391, so y = 17; then x = 159 − 7 × 17 = 40 (2)
  3. Fixed charge Rs 40, Rs 17 per kilometre; amount for 18 km = Rs 346 (1)
5
33Sn = n/2 [2a + (n − 1)d] with a = 220, d = 45: n/2 [440 + (n − 1) × 45] = 7175, that is, 9n2 + 79n − 2870 = 0, so n = 14 or n = −205/9. The number of months must be a natural number, so n = −205/9 is rejected; n = 14 months. The 14th term is 220 + 13 × 45 = 805, so Rs 805 for the last month.
  1. Forming the equation: Sn = n/2 [2a + (n − 1)d] with a = 220, d = 45: n/2 [440 + (n − 1) × 45] = 7175, that is, 9n2 + 79n − 2870 = 0 (2)
  2. Solving: n = 14 or n = −205/9 (2)
  3. Rejecting n = −205/9: 14 months; last term Rs 805 (1)
5
34Join ZW, meeting LG at B. In △ZDW, LB ∥ DW, so ZL/LD = ZB/BW (BPT). In △WZT, BG ∥ ZT, so WB/BZ = WG/GT, that is ZB/BW = TG/GW. Hence ZL/LD = TG/GW. TG = 9 cm and GW = 15 cm.
  1. Construction: join ZW meeting LG at B (1)
  2. BPT in △ZDW (1)
  3. BPT in △WZT and conclusion (1)
  4. TG = 9 cm (1)
  5. GW = 15 cm (1)
5
35Distance travelled = 66(√3 − 1) m ≈ 48.2 m; distance at the second observation = 66 m
  1. Angles of depression equal the angles of elevation from the ship (1)
  2. First distance = 66√3 m (1)
  3. Second distance = 66 m (1)
  4. Distance travelled = 66(√3 − 1) m (1)
  5. ≈ 48.2 m (√3 = 1.73) (1)
5
35 ORHeight of the chimney = 24(√3 + 1) m ≈ 65.5 m; distance = 24√3 m ≈ 41.5 m
  1. Correct figure described: horizontal line from the top of the building (1)
  2. Distance = 24√3 m from the angle of depression (1)
  3. Part above the roof = 24√3 m (1)
  4. Height = 24(√3 + 1) m (1)
  5. Values to one decimal (√3 = 1.73) (1)
5
Section E: 3 case-based questions of 4 marks each
36 (i)13 km
  1. √((−12)2 + (−5)2) = √169 — 13 km (1)
1
36 (ii)(−12.5, −4)
  1. ((−10 + (−15))/2, (−9 + 1)/2) — (−12.5, −4) (1)
1
36 (iii)HS + SM = 2√5 + 3√5 = 5√5 = HM, so H, S and M are collinear
  1. HS = √((−12 − (−10))2 + (−5 − (−9))2) = 2√5, SM = 3√5, HM = 5√5 km (1)
  2. HS + SM = 5√5 = HM, so the points are collinear (1)
2
36 (iii) ORRatio 9 : 1; point (−14.5, 0)
  1. Let the ratio be k : 1; the y-coordinate (k × 1 + (−9))/(k + 1) = 0 gives k = 9 (1)
  2. Ratio 9 : 1; x = (9 × (−15) + 1 × (−10))/10 = −14.5, so the point is (−14.5, 0) (1)
2
37 (i)
  1. 10√3 m (1)
1
37 (ii)
  1. 10 m (1)
1
37 (iii)40 m
  1. Height = 30 + 10 (1)
  2. 40 m (1)
2
37 (iii) OR20√3 m ≈ 34.6 m
  1. sin 60° = 30/s (1)
  2. 20√3 m ≈ 34.6 m (1)
2
38 (i)50 cm
  1. 92 − 2 × 21 — 50 cm (1)
1
38 (ii)6,600 cm2
  1. 2πrh = 2 × 22/7 × 21 × 50 — 6,600 cm2 (1)
1
38 (iii)12,144 cm2
  1. TSA = 2πrh + 2 × 2πr2 = 6,600 + 5,544 (1)
  2. = 12,144 cm2 (1)
2
38 (iii) OR108,108 cm3
  1. Volume = πr2h + (4/3)πr3 = 69,300 + 38,808 (1)
  2. = 108,108 cm3 (1)
2